No, this scenario is counter intuitive.
The model is that a card is picked at random, and then a side is displayed at random.
There are six equally probable results of this -- one for each face. Three of these result in displaying a green side and only one of those will have red on the other side.
The probability that the other side is also green given that a green side is displayed is: $2/3$.
Remark The information that "a green side is displayed" is not the same as knowing "the card has a green side", or even that "the card isn't double red."
By listing all the outcomes of the card drawn and side displayed (first or second).
$$\{\overbrace{\underbrace{GG:1, GG:2, GR:1}_\text{you see a green side}, GR:2}^\text{the card isn't double red}, RR:1, RR:2\}$$
The likelihood of drawing each card is equal, but if a card with two green sides is drawn it is twice as likely to display a green side than if the card with one green side were drawn. Therefore, if a green side is displayed the card is twice as likely to be a double green card than a single green card.
Let $C$ be the card drawn, taking values $\{0, 1, 2\}$ for the number of green sides. (So $C=0$ is the red-red card, et cetera).
Let $D$ be the colour displayed, taking values $\{0, 1\}$ for red or green respectively.
So we want to find: $\mathsf P(C=2 \mid D=1)$
$$\begin{align}
& = \cfrac{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)}{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)+\mathsf P(C=1)\cdot\mathsf P(D=1\mid C=1)+\mathsf P(C=0)\cdot\mathsf P(D=1\mid C=0)}
\\[1ex] & = \cfrac{\cfrac 1 3 \cdot\cfrac{2}{2}}{\cfrac 1 3 \cdot\cfrac{2}{2}+\cfrac 1 3 \cdot\cfrac{1}{2}+\cfrac 1 3 \cdot\cfrac{0}{2}}
\\[1ex] & = \dfrac 2 3
\end{align}$$
If you randomly decide upon n pairs of gloves, you will be satisfied if the set of left glove colours exactly matches the set of right glove colours. There are 10-choose-n sets of colours of size n, and since each is equally likely, there's a one in 10-choose-n chance of you being satisfied. 10-choose-n is 10!/(n!(10-n)!), so 1/10 for one glove, 1/45 for two, 1/120 for three etc.
If, on the other hand, you specifically search for a set of gloves that will satisfy you, the odds of being satisfied are obviously much higher. Rather than counting ways in which you might be satisfied, it's easier to identify the specific permutations where you will not be satisfied. As a commenter has already stated, it's equivalent to asking what proportion of permutations of 10 things have a cycle of length 4 or less. Think of it this way, you start with one pair of gloves, consider the left glove, and look for the pair with the matching right glove. Keep repeating this until the pair you're looking for next is the pair you started with. You now have a cycle in the permutation, and if it's of length four or less you've found a set of gloves to satisfy you.
Firstly observe that all permutations consist of a disjoint set of cycles. Which means, if a permutation has a cycle of length 8, then the remaining two elements must either be in a cycle of length 2 or two cycles of length 1. So, you will be satisfied if there are any cycles of length 1, 2, 3 or 4, and you will also be satisfied if there are any cycles of length 6, 7, 8 or 9 since all of those force there to be a cycle of length 4 or less in the remaining elements. The only options under which you will be unsatisfied are two cycles of length 5, or one cycle of length 10.
There are:
- 10! permutations in total
- 9! cycles of length 10, thus a 9!/10! = 1/10 chance of this happening
- 10-choose-5.4!.4! / 2 = 10!4!4!/(5!5!2) pairs of cycles of length 5, thus a 1/50 chance of this happening. That's 10-choose-5 choices for the members of the "first" of the cycles, 4! different 5-cycles for each of the two, and dividing by 2 because (as you observed) this counts each pair twice.
Final probability is 1 - 1/10 - 1/50 = 88% of being able to satisfy yourself.
Best Answer
We probably need to divide into cases. The following "grinding things out" way is not elegant, but it works. There are $4$ cases, a lot, but they come in very similar pairs.
We count the number of "good" choices of $3$ gloves.
(i) Three green. There are $\binom{4}{3}$ ways to do this, all good, since all such choices lead to a matching pair.
(ii) Three red. There are $\binom{6}{3}$ ways to choose $3$ red. Of them, all but $2$ (all left and all right) are good. So the number of good of this type is $\binom{6}{3}-2$.
(iii) Two green, one red. We must pick one left green and one right green, which can be done in $\binom{2}{1}\binom{2}{1}$ ways. For each such way, we can pick any red. That gives $\binom{2}{1}\binom{2}{1}\binom{6}{1}$ good.
(iv) Two red, one green. The same reasoning as the one above gives $\binom{3}{1}\binom{3}{1}\binom{4}{1}$.